Download Biostatistical Methods and more Exams Biostatistics in PDF only on Docsity!
Biostatistical Methods I PUBH 7150/ Final Exam Summer 2022 Instructions
Please answer all questions thoroughly and show all work. Please save document with the
following title: Firstname_Lastname_PUBH7150_Midterm and upload once you are done. The
point values out of 100 are provided in each section. You have 3.5 hours to complete the exam.
Course resources are permitted (open note and open book) with the exception of another
person. No other person or resources other than course material is allowed to assist you on
this exam. Please keep the copy of your exam and answers for the exam only by yourself.
Please make sure do not share with anyone else.
For all the hypothesis testing question, please provide all the five steps. Time and Date Started: Time Finished: I worked on the exam independently and did not seeking any assists from other person ()
Grading Rubric For Master students: Total Points Points Earned Multiple Choice 20 Problem 1 15 Problem 2 25 Problem 3 40 Extra Credit Question 10 Total 100
_ b ___ 6. If a Spearman correlation coefficient is r = 0.8 then
a) At least one of the variables are not normally distributed.
b) The relationship between the two variables is positive.
c) The coefficient of determination is 64%
d) Both variables are numeric and continuous.
_ a __ _ __ 7. Which panel below shows the strongest positive correlation: __ c ____8. Suppose we are comparing the means of 5 different groups and have rejected the overall F test suggesting that at least one mean is different. Our next step is to use multiple comparisons to test which means are different. Which is the correct pairwise comparison p-value using Bonferroni correction when ฮฑ=0.05? a) 0. b) 0. c) 0. d) 1. _ d _____9. Which one is the appropriate approach for testing the equality of two proportions when we have a very large sample size? a) t-test. b) Sign test. c) one-way ANOVA procedure. d) z-test __ c ___ 10. What is the probability of testing positive on a screening test for a disease, given you truly have the disease? a) Specificity b) Accuracy c) Sensitivity d) Prevalence
Question 1 (15 points)
We are conducting a study to determine the effects of drug on hypertension. The arterial blood
pressure of a random sample of 12 patients is measured before and after the treatment, the
data are showing below:
Patient ID Before After Diff Sign
Please conduct a sign test to find out whether the surgery changed the arterial blood pressure.
- Hypothesis: H0: MD difference โค 0, H1: MD difference > 0
- Critical value: Using sign test critical value table, With n = (number of +) + (number of -) = 7+2 and ฮฑ =0.05, Xcrit = 1
- Test Statistic: Xtest = min [(number of +), (number of -)] = min [7, 2] = 2 p= 0.5, n= 9, x= 2 P(k โค x|n,p) = (^) โ k = 0 x
n!
k! (^ n โ k )^!
p
k
( 1 โ p )
n โ k P(k โค 2|9,0.5)= (^) โ k = 0 9
k! (^9 โ k )^!
0.5 k^ ( 1 โ0.5)
9 โ k
0
9
P(k โค 2|9,0.5)= 0.
2 x 0.08984= 0.
- Decision: (Xtest = 2) > Xcrit = 1) So do not reject H 0
b) Please calculate the odds ratio of having disease B comparing to no disease group under the chemical A exposure and its 95% confidence interval Exposure Total chemical A exposure No chemical A exposure Disease Disease B 125 50 175 No Disease B 75 150 225 Total 200 200 400 Odds Ratio = OR = (125150)/(7550) = 5 95% CI for OR = exp[ln(OR) ยฑ Zcritical * sqrt ( (1/n11)+(1/n12)+(1/n21)+(1/n22)] = exp[1.609 ยฑ 1.96 * sqrt((1/125)+(1/50)+(1/75)+(1/150))] = exp[1.609 ยฑ 1.96 * 0.219] = exp[1.609 ยฑ 0.429] L = exp [1.609-0.429] = 3. U = exp [1.609+0.429] = 7. The 95% CI that the odds of disease for persons exposed to the mysterious chemical A are between 3.254 and 7.675 times that of persons not exposed.
Question 2 (40 points) Researchers are looking to compare two different drug interventions. The patientโs addiction scores for the two different interventions group are shown as the following table. Column 3 represents the susceptibility measures for patients in group 1. (Assume data from all the three columns are normally distributed). Score Group 1 Score Group 2 Susceptibility Group 5 4 3. 3 3 2. 1 4 -4. 2 5 -4. 3 4 -1. 7 1 2. 6 6 -1. 5 3 4. 2 7 -0. 4 6 -2. a) Calculate the mean, median and model for addiction scores in both groups. (8 pts) For scores in group 1: Mean = (5+3+1+2+3+7+6+5+2+4) / 10 = 38/10 = 3. The data in order is as: 1,2,2,3,3,4,5,5,6, Median = (3+4)/2 = 3. Mode = most frequent observation = 2,3,5 (multiple modes) For scores in group 2: Mean = (4+3+4+5+4+1+6+3+7+6) / 10 = 43/10 = 4. The data in order is as: 1,3,3,4,4,4,5,6,6, Median = (4+4)/2 = 4 Mode = most frequent observation = 4
c) Test if group 2 mean addiction score is higher compared to group 1 mean addiction score. In the essence of time, we calculated the folded F-test for you and the associated p-value from the folded F-test is 0.0525. (8 pts)
- Hypothesis: ๐ป0: ๐1 = ๐ 2 ๐ป1: ๐1 < ๐ 2 n1 = 10, n2=10, the degrees of freedom = (10-1) + (10-1) = 9 + 9 = 18 Let ฮฑ = 0.
- Test Statistic:
t =
x 1 โ x 2
SE
x 1 โ x 2 โ
1 n 1 +^ 1 n 1
x 1 =3.8 , x 2 =4.
s 1 =1.932 , s 2 =ยฟ1.
SEx 1 โ x 2 =
( n 1 โ^1 ) s 1
2 +( n
2 โ^1 )^ s 2
2
n 1 + n 2 โ 2
t =
- Critical value of t At ฮฑ =0.05 and degrees of freedom = 18, t critical = 2.
- Conclusion: Since the computed t-statistic is less than the t-critical, fail to reject null hypothesis. Hence, group 2 mean addiction score is not significantly higher compared to group 1 mean addiction score
- Calculate the correlation between group 1 addiction score and patient susceptibility. Comment on the strength and the direction of the correlation. (8 pts) Score Group 1 (X) Susceptibility (group 1) (y) (x-xbar) (x-xbar)2 (y-ybar) (y-ybar)2 (x-xbar) * (y-ybar) 5 3.56 1.2 1.44 3.933 15.468 4. 3 2.23 -0.8 0.64 2.603 6.776 -2. 1 -4.93 -2.8 7.84 -4.557 20.766 12. 2 -4.97 -1.8 3.24 -4.597 21.132 (^) 8. 3 -1.92 -0.8 0.64 -1.547 2.393 (^) 1. 7 2.85 3.2 10.24 3.223 10.388 (^) 10. 6 -1.98 2.2 4.84 -1.607 2.582 (^) -3. 5 4.35 1.2 1.44 4.723 22.307 (^) 5. 2 -0.44 -1.8 3.24 -0.067 0.004 0. 4 -2.48 0.2 0.04 -2.107 4.439 -0.
r =
โ ( xi โ x^ )โ(^ yi โ^ y^ ) โโ (^ xi โ x ) 2 โ (^ yi โ^ y )
2 =^
โ33.6โ106.^
Correlation between group 1 addiction score and patient susceptibility is 0.62, which is high degree positive correlation.
Bonus Question (10 pts)
1. A set of data is known to follow a normal distribution N(ฮผ, ฯ^2 ), that is, mean of ฮผ, and
variance of ฯ^2. If we take a sample of 100 and use the Z table to construct the
confidence interval for ฮผ, the result is (99.51,100.49). What is ฯ^2? (5 pts)
95 % CI = x ยฑ
โ n
Given that 95% CI is (99.51,100.49).
Hence
x โ
โ n
x โ
10 โ x โ1.96โ ฯ =99.51 ----- e
10 * x^ +1.96โ ฯ =100.49^ ----- e
By Solving e1 and e2, we get ฯ^ =0.25, Hence, ฯ^2 =0.
2. The data is from a study on treatments for healing severe infections. Randomly assigned
test treatment and control are compared to determine whether the rates of favorable
response are the same. Please conduct the appropriate analysis. (5 pts)
Severe Infection Treatment Outcomes
Treatment Favorable Unfavorable Total
Test 10 (a) 2 (b) 12 (a+b)
Control 2 (c) 4 (d) 6 (c+d)
Total 12 (a+c) 6 (b+d) 18
H 0 : Severe Infection Treatment Outcome is independent of treatment
H 1 : Severe Infection Treatment Outcome is not independent of treatment
p= [ ( a + b )! ( c + d )! ( a + c )! ( b + d )! ) / a! b! c! d! N! = [12! * 6! * 12! * 6!] / (10!2! 2!* 4!) = 0.** Hence there is no evidence to support H1. Conclusion:
Severe Infection Treatment Outcome is independent of treatment
